# Proof that $2^n \times 2^n$ operator be decomposed in terms of $2 \times 2$ operators

What is the proof that any $$2^n\times 2^n$$ quantum operator can be expressed in terms of the tensor product of $$n$$ number of $$2\times 2$$ quantum operators acting on a single qubit space each?

• The answer is indicating sums are allowed in addition to just tensor product. But additive decomposition vs multiplicative decomposition wasn't clear in the question. Could you edit the question for if/if not sums are allowed – AHusain Jan 31 '19 at 19:40
• Sums are allowed. Sorry that isn't clear in the question. I will edit it. – Siddhant Singh Jan 31 '19 at 20:19

I presume you mean a $$2^n\times 2^n$$ quantum operator, $$U$$?
Let's assume we can write $$U=\sum_{x,y\in\{0,1\}^n}U_{xy}|x\rangle\langle y|.$$ All we have to do is show that we can construct any $$|x\rangle\langle y|$$ using Pauli operators. But this is just the same as $$\bigotimes_{i=1}^n|x_i\rangle\langle y_i|,$$ so provided I can create any $$|x_i\rangle\langle y_i|$$ for $$x_i,y_i\in\{0,1\}$$ using Pauli matrices, I'm done. It's a simple exercise to check $$|0\rangle\langle 0|=(\mathbb{I}+Z)/2\qquad |1\rangle\langle 1|=(\mathbb{I}-Z)/2$$ and $$|0\rangle\langle 1|=(X+iY)/2\qquad |1\rangle\langle 0|=(X-iY)/2.$$